prove-0-1-x-x-2-1-x-ln-x-dx-ln-4-pi-

Question Number 165152 by mnjuly1970 last updated on 26/Jan/22

p r o v e

Ω = \int_{0}^{1} \frac{x - x^{2}}{(1 + x) l n (x)} d x = l n (\frac{4}{π})

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Answered by mindispower last updated on 26/Jan/22

f (a) = \int_{0}^{1} \frac{x - x^{a + 1}}{(1 + x) l n (x)} d x, f (0) = 0

Ω m u s t b e e < 0 l n (\frac{4}{π}) > 0

f^{'} (a) = \int_{0}^{1} \frac{- x^{1 + a}}{1 + x}

= \int_{0}^{1} \frac{x^{2 + a} - x^{1 + a}}{1 - x^{2}} d x

= \frac{1}{2} \int_{0}^{1} \frac{t^{\frac{a + 1}{2}} - t^{\frac{a}{2}}}{1 - t} . d t

r e c a l l Ψ (z + 1) = - γ + \int_{0}^{1} \frac{1 - x^{z}}{1 - x} d x

f^{^{'}} (a) = \frac{1}{2} {Ψ (\frac{a}{2} + 1) - Ψ (\frac{a + 3}{2})}

f (a) = l n (\frac{Γ (\frac{a + 2}{2})}{Γ (\frac{a + 3}{2})}) + c

f (0) = 0 \Rightarrow c = - l n (\frac{2}{\sqrt{π}}) = l n (\frac{\sqrt{π}}{2})

Ω = f (1) = l n (\frac{\frac{\sqrt{π}}{2}}{1}) = l n (\frac{\sqrt{π}}{2}) + l n (\frac{\sqrt{π}}{2}) = l n (\frac{π}{4})

Ω = l n (\frac{π}{4})

Commented by mnjuly1970 last updated on 27/Jan/22

t h a n k s a l o t s i r p o w e r . . g r a t e f u l

Commented by mindispower last updated on 27/Jan/22

w t h e P l e a s u r H a v e a n i c e D a y